What is the d/(dt)((s)(4t^3-3t^4)) ?

The d/(dt)((s)(4t^3-3t^4)) is s(12t^2-12t^3)

What is the first d/(dt)((s)(4t^3-3t^4)) ?

The first d/(dt)((s)(4t^3-3t^4)) is s(12t^2-12t^3)

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\frac{d}{d t} ((s) (4 t^{3} - 3 t^{4}))

s (12 t^{2} - 12 t^{3})

Solution steps

\frac{d}{d t} ((s) (4 t^{3} - 3 t^{4}))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= s \frac{d}{d t} (4 t^{3} - 3 t^{4})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= s (\frac{d}{d t} (4 t^{3}) - \frac{d}{d t} (3 t^{4}))

\frac{d}{d t} (4 t^{3}) = 12 t^{2}

\frac{d}{d t} (3 t^{4}) = 12 t^{3}

= s (12 t^{2} - 12 t^{3})

\frac{( 5 - y ) ^{2}}{100}

\frac{1}{( s ( s ^{2} + 4 ) )}

\frac{e ^{x}}{x ^{2} \cdot ( x + 1 )}

x (1 + e^{38.6817400000000 \dots \cdot x}) - 2.5

\frac{2 x - 4}{( x + 1 ) ( x + 3 )}

What is the d/(dt)((s)(4t^3-3t^4)) ?
The d/(dt)((s)(4t^3-3t^4)) is s(12t^2-12t^3)
What is the first d/(dt)((s)(4t^3-3t^4)) ?
The first d/(dt)((s)(4t^3-3t^4)) is s(12t^2-12t^3)